Issue 28

A. Brotzu et alii, Frattura ed Integrità Strutturale, 28 (2014) 19-31 ; DOI: 10.3221/IGF-ESIS.28.03 29 the real one, twopathswere followed. The first consists in applying a load of 1 N on the end of the instrument perpendicularly to the axis of the instrument itself. This load was assessed by flexion test, performed by a tensile machine, blocking the cutter with a straight canal of length equal to that of the straight stretch of the simulacrum and applying a displacement of the tip 3mm in the direction perpendicular to the axis of the cutter. The applied load vs. tipdisplacements has been recorded. The second, instead, consists in inserting the tip and imposing a 3 mm failure of the bond. The two simulations gave similar results with the development of a stress state characterized by fibers stretched in the upper part of the instrument and compressed fibers in the lower part. The displacement of the tip of the instrument obtained in the first simulation is slightly less than the desired 3mm. Fig. 24 and 25 show the results of the two simulations expressed as stresses calculated assuming a homogeneousmaterial with 40GPa Youngmodulus. Themaximum values are detected near the areawhere it begins the curvature of the instrument. Tab. 5 shows the calculated maximum tensile and compressive strains. In the simulation with load applied at the tip a  of 4.1% is obtained, while with the displacement imposed  is significantly higher and amounted to 7.4%. TensileMaximum strain CompressionMaximum strain Tip applied load (1N) 2.3% -1.8% TipConstrain failure 4% -3.4% Table 5 : Calculatedmaximum tensile and compressive strains. Figure 24 : FEM simulationwith theTip applied load. Figure 25 : FEM simulationof theTIP constrain failure. C ONCLUSIONS he study experimental evidences are as follows:  The considered rotary instruments fail in twodifferent ways depending on the operatingmode. The samples testedwithout lubricant often fail by unwinding (high irreversible plastic deformation). The failure takes place T